The magnitude of acceleration of the electron in the $n$th orbit of a hydrogen atom is $a_{H}$ and that of a singly ionised helium atom is $a_{He}$. The ratio of $a_{H} : a_{He}$ is

Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom,the photon emitted in the $n = 2$ to $n = 1$ transition has energy $74.8 \ eV$ higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionization energy of the hydrogen atom is $13.6 \ eV$. The value of $Z$ is:

$A$ particle of mass $200 \, MeV/c^2$ collides with a hydrogen atom at rest. Soon after the collision,the particle comes to rest,and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in $eV$) is $\frac{N}{4}$. The value of $N$ is: (Given the mass of the hydrogen atom to be $1 \, GeV/c^2$)

According to Bohr's model, the highest kinetic energy is associated with the electron in the

Assuming the atom is in the ground state, the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is [$\mu_0 =$ permeability of free space, $m =$ mass of electron, $\epsilon_0 =$ permittivity of free space, $h =$ Planck's constant].

In a Bohr atom,the total energy of an electron in the $n$-th allowed orbit is $E_n$ and its angular momentum is $J_n$. Then: